Question

An aqueous solution contains 0.483 M ammonia (NH₃).

How many mL of 0.206 M hydroiodic acid would have to be added to 150 mL of this solution in order to prepare a buffer with a pH of 9.090?

mL

Answer 1

concentration of NH₃ = [NH₂] = 0.483 M = M, Volume of NH₃ = v= 150ml. milimoles of NH3 = M, V, = 0.483 x 150 = 72.45 mmol Concentration of hI = 0.206 M = M₂ Volume of HI = ? = V₂ milimoles of HI = M, V, = 0.206 Ng why + HI - NHYT I" 92.45 o initial Mooles 0. 206 final (12:45 – 0.20602) (0-2069,-0.2007) mmoles 0 pkp of NH₃ = 4.75 PH = 8090 pon = 14-PH2 14-googo = 4.91 pour рең = ph, + IP | С вичі) Cang]

40 91 = 4.75 + log / 0.206 V2 (12:45- 0.2064 1.445= 0.2067₂ 72.45 - 012062 V = 207.85 ml Volume of HI should be added - (207.85 ml